PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 138, Number 2, February 2010, Pages 593–603 S 0002-9939(09)10157-0 Article electronically published on September 25, 2009 THE LAPLACE TRANSFORM OF THE PSI FUNCTION ATUL DIXIT (Communicated by Peter A. Clarkson) Abstract. An expression for the Laplace transform of the psi function ∞ L(a):= e−atψ(t +1)dt 0 is derived using two different methods. It is then applied to evaluate the definite integral 4 ∞ x2 dx M(a)= , 2 2 −a π 0 x +ln (2e cos x) for a>ln 2 and to resolve a conjecture posed by Olivier Oloa. 1. Introduction Let ψ(x) denote the logarithmic derivative of the gamma function Γ(x), i.e., Γ (x) (1.1) ψ(x)= . Γ(x) The psi function has been studied extensively and still continues to receive attention from many mathematicians. Many of its properties are listed in [6, pp. 952–955]. Surprisingly, an explicit formula for the Laplace transform of the psi function, i.e., ∞ (1.2) L(a):= e−atψ(t +1)dt, 0 is absent from the literature. Recently in [5], the nature of the Laplace trans- form was studied by demonstrating the relationship between L(a) and the Glasser- Manna-Oloa integral 4 ∞ x2 dx (1.3) M(a):= , 2 2 −a π 0 x +ln (2e cos x) namely, that, for a>ln 2, γ (1.4) M(a)=L(a)+ , a where γ is the Euler’s constant. In [1], T. Amdeberhan, O. Espinosa and V. H. Moll obtained certain analytic expressions for M(a) in the complementary range 0 <a≤ ln 2. Our goal here is to derive an explicit expression of L(a)fora>0intermsof elementary functions and a certain simple infinite series which cannot be evaluated Received by the editors October 28, 2008. 2000 Mathematics Subject Classification. Primary 33B15. Key words and phrases. Laplace transform, gamma function, psi function. c 2009 American Mathematical Society Reverts to public domain 28 years from publication 593 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 594 ATUL DIXIT in terms of elementary functions. We give two different proofs, the second of which is shorter. However, it makes use of a formula (equation 2.31) in [1], viz., equation (3.1) in our paper, which requires considerable work for its derivation. Also, our first proof is self-contained and elementary. We then use our expression for the Laplace transform of the psi function to evaluate M(a)fora>ln 2 and thereby resolve a conjecture posed by Olivier Oloa in [8]. Our main theorem can be stated as follows: Theorem 1.1. Let a>0.Ifψ(x) is defined as in (1.1),then ∞ ∞ 1 1 2π ln n L(a)= e−atψ(t +1)dt = − +1 ln +2a a − 2 2 2 0 e 1 a a a +4n π n=1 1 ia ia γ +lna (1.5) + ψ + ψ − − . 4 2π 2π a 2. Laplace transform of the psi function: First proof First, from [2, p. 259, formula 6.3.18] for | arg z| <π,asz →∞, 1 1 1 1 (2.1) ψ(z) ∼ ln z − − + − + ··· . 2z 12z2 120z4 252z6 We now state a lemma1 which will be subsequently used in the proof. The integral involved in the lemma exists because of (2.1). This integral gives the motivation for decomposing L(a) into the two integrals mentioned in equation (2.9) below. This integral evaluation is implicit in the work of A.P. Guinand on a certain transfor- mation formula involving the psi function (see the footnote on the last page in [7]), but he neither proves it nor explicitly states it in [7]. Lemma 2.1. ∞ 1 1 (2.2) ψ(t +1)− − ln t dt = ln 2π. 0 2(t +1) 2 Proof. Let I denote the integral on the left-hand side. Then, ∞ d etΓ(t +1) I = ln √ dt t 0 dt t t +1 etΓ(t +1) etΓ(t +1) = lim ln √ − lim ln √ t→∞ tt t +1 t→0 tt t +1 etΓ(t +1) 1 =ln lim √ − ln lim etΓ(t +1) − lim t ln t − lim ln(t +1) t→∞ tt t +1 t→0 t→0 t→0 2 etΓ(t +1) (2.3) = ln lim √ . t→∞ tt t +1 Stirling’s formula [6, p. 945, formula 8.327] tells us that √ (2.4) Γ(z) ∼ zz−1/2e−z 2π, as |z|→∞and | arg z|≤π − δ,where0<δ<π. Hence, employing (2.4), we find that √ etΓ(t +1) 1 t 2π (2.5) √ ∼ 1+ , tt t +1 t e 1The author is indebted to M. L. Glasser for the proof of this lemma. The author’s original proof is significantly longer than that given by Glasser. License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use THE LAPLACE TRANSFORM OF THE PSI FUNCTION 595 so that etΓ(t +1) √ (2.6) lim √ = 2π. t→∞ tt t +1 Thus from (2.3) and (2.6), we conclude that 1 (2.7) I = ln 2π. 2 Now we begin with the proof of Theorem 1.1. Without mentioning it explicitly, throughout this paper we make use of the fact that 1 (2.8) ψ(x +1)=ψ(x)+ . x Using (2.1), we can decompose the Laplace transform of the psi function into two integrals, ∞ ∞ 1 e−atψ(t +1)dt = e−at ψ(t +1)− − ln t dt 2(t +1) 0 0 ∞ 1 + e−at +lnt dt 0 2(t +1) (2.9) = I1 + I2, say, where (2.1) ensures that I1 exists. The integral I2 is easy to evaluate. Using the definition of the incomplete gamma function ∞ e−t (2.10) Γ(0,a)= dt, a t and the fact [6, p. 602, formula 4.331, no. 1] that ∞ γ +lna (2.11) e−ax ln xdx= − , for a>0, 0 a we obtain ea γ +lna (2.12) I = Γ(0,a) − . 2 2 a Now let I1 = I1(a). Differentiating under the integral sign, we have ∞ 1 I (a)= −te−at ψ(t +1)− − ln t dt 1 2(t +1) 0 ∞ t = e−at t ln t + − tψ(t +1) dt 2(t +1) 0 ∞ 1 1 1 = e−at t ln t + − − t ψ(t)+ dt 2 2(t +1) t 0 ∞ e−at ∞ 1 = − dt − te−at ψ(t)+ − ln t dt 0 2(t +1) 0 2t ea (2.13) = − Γ(0,a) − I , say. 2 3 Using a well-known formula [9, p. 248] in a slightly simplified form, we have for Re z>0, 1 ∞ 1 1 1 (2.14) ψ(z)+ − ln z = − − + e−tz dt. t − 2z 0 e 1 t 2 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 596 ATUL DIXIT Thus, ∞ ∞ 1 1 1 I = − te−at − + e−xt dx dt 3 ex − 1 x 2 0 0 ∞ 1 1 1 ∞ = − − + te−(x+a)t dt dx ex − 1 x 2 0 0 ∞ 1 1 1 dx (2.15) = − − + . x − 2 0 e 1 x 2 (x + a) The inversion of the order of integration above is easily justifiable. Next from [3, p. 191], we have for x =0, ∞ 1 1 1 1 1 (2.16) = − + . x2 +4n2π2 2x ex − 1 x 2 n=1 Hence, ∞ ∞ 2xdx 1 I = − 3 (x + a)2 x2 +4n2π2 0 n=1 ∞ 1 ∞ 2ax = − dx a (x + a)2(x2 +4n2π2) n=1 0 ∞ 1 ∞ (x2 +2ax + a2) − (x2 + a2) = − dx a (x + a)2(x2 +4n2π2) n=1 0 ∞ 1 ∞ dx ∞ x2 + a2 (2.17) = − − dx , a x2 +4n2π2 (x + a)2(x2 +4n2π2) n=1 0 0 where the inversion of integration and summation can be easily seen via absolute convergence. Using Mathematica, we find that ∞ x2 + a2 dx 2 2 2 2 0 (x + a) (x +4n π ) a4 − 8a2n2π2 +32an3π2 +16n4π4 +8a3n(1 + ln(2)) + 32an3π2 ln(a/2nπ)+8a3n ln(nπ/a) = 4n(a2 +4n2π2)2 (a2 +4n2π2)2 − 16a2n2π2 +8a3n(1 + ln(2πn/a)) + 32an3π2(1 − ln(2πn/a)) = . 4n(a2 +4n2π2)2 (2.18) Hence from (2.17) and (2.18), we deduce after some simplification that ∞ 1 32an3π2(1 − ln(2πn/a)) + 8a3n(1 + ln(2πn/a)) − 16a2n2π2 I = 3 4a n(a2 +4n2π2)2 n=1 (2.19) = S1 + S2 + S3 + S4 − S5, License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use THE LAPLACE TRANSFORM OF THE PSI FUNCTION 597 where ∞ n2 (2.20) S =8π2 , 1 (a2 +4n2π2)2 n=1 ∞ ln(2π/a) 8a3n − 32an3π2 (2.21) S = , 2 4a n(a2 +4n2π2)2 n=1 ∞ 1 (8a3n − 32an3π2)lnn (2.22) S = , 3 4a n(a2 +4n2π2)2 n=1 ∞ 1 (2.23) S =2a2 , 4 (a2 +4n2π2)2 n=1 and ∞ n (2.24) S =4aπ2 . 5 (a2 +4n2π2)2 n=1 Decomposing S1 into partial fractions, ∞ ∞ 1 1 S =2 − 2a2 1 a2 +4n2π2 (a2 +4n2π2)2 n=1 n=1 ∞ ∞ 1 d 1 =2 + a a2 +4n2π2 da a2 +4n2π2 n=1 n=1 ∞ ∞ 1 d 1 (2.25) =2 + a , a2 +4n2π2 da a2 +4n2π2 n=1 n=1 and then using (2.16), we find after some simplification that 2coth(a/2) − a/ sinh2(a/2) (2.26) S = . 1 8a Similarly, breaking S2 into partial fractions and then employing (2.16) and (2.26), we find upon simplification that ln(2π/a) a 4 (2.27) S2 = − .